Why is it useful to classify the vector bundles of a space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:00:08Z http://mathoverflow.net/feeds/question/16240 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space Why is it useful to classify the vector bundles of a space? foocorge 2010-02-24T07:40:28Z 2010-02-24T20:23:24Z <p>It seems to me that vector bundles are useful because they allow us to bring to bear all of the linear algebra we know to aid in the study of topological spaces. Now, I've read somewhere that it is an important and difficult problem to classify all of the vector bundles of a space. I'm willing to accept that the problem is difficult, but why is it important? What are some applications of such a classification?</p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16241#16241 Answer by Steve Huntsman for Why is it useful to classify the vector bundles of a space? Steve Huntsman 2010-02-24T07:49:02Z 2010-02-24T14:39:15Z <p><a href="http://en.wikipedia.org/wiki/Topological_K-theory" rel="nofollow">Topological K-theory</a> is quite useful: for example, it played a <a href="http://en.wikipedia.org/wiki/Atiyah%E2%80%93Singer_index_theorem#K_theory" rel="nofollow">key role</a> in understanding the <a href="http://en.wikipedia.org/wiki/Atiyah%E2%80%93Singer_index_theorem" rel="nofollow">index theorem</a>.</p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16244#16244 Answer by Ryan Budney for Why is it useful to classify the vector bundles of a space? Ryan Budney 2010-02-24T08:10:47Z 2010-02-24T08:10:47Z <p>If your space is a manifold, knowing the vector bundles over that space amounts to knowing all of its <a href="http://en.wikipedia.org/wiki/Tubular_neighborhood" rel="nofollow">tubular neighbourhoods</a> when you embed the space in another manifold. This frequently allows you to find many relationships between the two manifolds.</p> <p>One classical application would be the proof that all smooth embeddings $S^n \to S^{n+2}$ (co-dimension two embedding of a sphere in a sphere) has a Seifert surface -- meaning there is an embedded, orientable $n+1$-manifold $M \to S^{n+2}$ whose boundary is the $n$-sphere. One of the main steps is showing that the $n$-sphere in $S^{n+2}$ has a trivial tubular neighbourhood. </p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16245#16245 Answer by Mikael Vejdemo-Johansson for Why is it useful to classify the vector bundles of a space? Mikael Vejdemo-Johansson 2010-02-24T08:16:41Z 2010-02-24T08:16:41Z <p>While not specific to vector bundles as such, classifying any kind of structure usually gives some sort of deeper understanding of the structure, and more importantly, tends to yield good and often compact descriptors.</p> <p>For instance, the classification of modules over (nice enough) algebras gives us all sorts of interesting invariants and decompositions: the classification of modules over PIDs gives us the Jordan normal form, with all the power it brings to linear algebra; and the classification of modules over quiver algebras (tame modules exist over a few families of Dynkin diagrams; all others admit wild modules) is what forms the basis for persistence diagrams - a formalization (and visualization) of the idea that classes in the homology of a filtered simplicial complex carry lifetimes that end up telling us much about the topology of the things we used to construct the simplicial complex in the first place.</p> <p>These ideas, again, are currently being applied in data analysis, in group cohomology and in computer graphics.</p> <p>All of which is to say that classification in general is good, and vector bundles are interesting, thus classifying vector bundles is good.</p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16248#16248 Answer by Pascal Lambrechts for Why is it useful to classify the vector bundles of a space? Pascal Lambrechts 2010-02-24T08:59:25Z 2010-02-24T08:59:25Z <p>I would say that the importance of the classification of vector bundles comes first from the fact that it leads naturally to the "characteristic classes" and their complete description. Characteristic classes are computable and powerful invariants of vector bundles. Look at the book Milnor-Stasheff "Characteristic classes" which is the wonderful classic in the domain. There you find first applications of the classification of vector bundles, the fundamental classification theorem being that isomorphism classes of rank n vector bundle E-->X over a space X are in 1-1 correspondance with homotopy classes of maps g:X-->BO(n) where BO(n) is an explicit topological space, namely it is the grasmannian of n-vector spaces inside R<sup>inifinity</sup>. The usage of this theorem is then that, using the cohomology of BO(n), you can build and classify very interesting and computable invariants of vector bundles (the more famous being the Euler class, the Stiefel-Whitney classes, the Pontrjagin classes and (for complex vector bundles) the Chern classes.) This is whar are called characteristic classes of vector bundles.</p> <p>Among applications of this given in Milnor-Stasheff are the following: <br>- you can prove that the real projective spaces P<sup>n</sup>(R) (which is an n-dimensional manifold) does not embed in R<sup>n+k</sup> for low values of k (depending on n). For example the 8-dimensional projective space does not embed in R<sup>14</sup>. <br> -you can prove that it is impossible to define a bilinear multiplication on R<sup>n</sup> without zero divisor (i.e. such that x.y&ne;0 when x&ne;0 and y&ne;0) unless n is a power of 2 (but when n=2 you have the complex multiplication, when n is 4 you have the quartenions, for n=8 you have octonions). This was latter improved by Adams who showed that n=1,2,4,8 are actyually the only possible values. <br> - you can classify compact smooth manifolds "up to cobordism" (Thom theorem). This was a first step to the classification of smooth manifolds up to diffeomorphism (surgery theory). The classification up to cobordism is very directly related to the cohomology of BO(n). <br> - another striking application which is not completely detailled in Milnor-Stasheff is the construction of exotic spheres, i.e. smooth manifolds homeomorphic but not diffeomorphic to a sphere (for example in dimension 7). A key ingredient in the proof is Hirzebruch signature formula which implies that some vector bundles cannot be the tangent bundle of a compact smooth manifold. <br> -extending this you can also build examples of topological manifolds which cannot have a smooth structure <br> - and this was only the beginning in the late 1950's :-). Then came K-theory, the complete classification of exotic spoheres, the classification of smooth manioflds, etcetera...</p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16252#16252 Answer by Daniel Moskovich for Why is it useful to classify the vector bundles of a space? Daniel Moskovich 2010-02-24T11:28:50Z 2010-02-24T11:28:50Z <p>This is a variation on Ryan's answer.<br> In PL topology, from which all of geometric topology sprang, every object X comes equipped with an embedding into R<sup>n</sup> for n sufficiently large. With the rise of homological algebra, and the shift of focus towards the smooth and topological categories, it became useful to separate between the chain complex of X over some ring (which you might use to calculate homology), which tells you something like how X is glued together out of simplexes, and between the information about how X might be locally embedded, which is the bundle theory over X. These two pieces of information about X are disjoint (although I'm now reading about quadratic structures on chain complexes, a formalism which combines them in some sense). In particular, the algebraic theory of chain complexes is a long way from the geometric topology of manifolds, even in the presence of Poincare duality structure.<br> All of this is part of the effort to view a space "intrinsically". Vector bundles are an abstraction of the tubular neighbourhood of X in some big Euclidean space, in the sense that chain complexes are an abstraction of stacking simplexes to build X. In this way, they become the natural tool to study "intrinsic external" structure in such settings as surgery problems, passage between categories (smooth, PL, and topological), differential topology over X, and so on.</p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16264#16264 Answer by Igor Belegradek for Why is it useful to classify the vector bundles of a space? Igor Belegradek 2010-02-24T14:38:47Z 2010-02-24T20:23:24Z <p>I am not sure why you think the classification of vector bundles is a difficult problem. It quickly reduces to homotopy theoretic issues that are quite well understood (as well as anything in homotopy theory can be understood).</p> <p>Vector bundles are important because they give natural invariants of manifolds (tangent bundle and its characteristic classes), and embeddings (normal bundle), as well as a natural place where various structures attached to a manifold live (e.g. metrics, connections, tensors).</p> <p>Whether the classification of bundles matters depends on what you do. To make an analogy if you are studying PDE on $\mathbb R^n$, which is a huge subject, then you might not care about classifications of manifolds at all; except that PDE do come in handy in classifying 3-manifolds. As such classification of vector bundles isn't used in most of the mathematics, rather it represents a basic result in topology. </p> <p>In fact I am curious how often do you use classification of vector bundles (be that identification of $\mathbb R^k$-bundles over $X$ with $[X, BO(k)]$ or the fact that complex line bundles are classified by first Chern class, or classification of bundles over complexes of dimension $\le 4$ in terms of characteristic classes)? In my own research I have used all this extensively but my impression that this is quite rare. </p> http://mathoverflow.net/questions/16240/why-is-it-useful-to-classify-the-vector-bundles-of-a-space/16288#16288 Answer by Ryan Budney for Why is it useful to classify the vector bundles of a space? Ryan Budney 2010-02-24T18:58:37Z 2010-02-24T18:58:37Z <p>I suppose a simpler answer to your question is that the classification of vector bundles is to vector bundles as Whitney's embedding theorem is to manifolds. </p> <p>Specifically, Whitney's weak embedding theorem says that all n-manifolds embed in $\mathbb R^{2n+1}$ and that any two embeddings of an $n$-manifold in $\mathbb R^{2n+2}$ are isotopic. So studying the "abstract" problem of $n$-manifolds up to diffeomorphism is equivalent to the "less abstract" problem of studying $n$-dimensional submanifolds of $\mathbb R^{2n+2}$ up to isotopy. </p> <p>Whitney's proof of the above fact is almost exactly how the proof of the classification of vector bundles works. Moreover, they're philosophically almost identical proofs, as the Whitney theorem says abstract manifolds are submanifolds. The classification theorem says abstract vector bundles can be thought of as having their fibres in Euclidean space. </p>